1: Ta có: \(\sqrt{4x^2-12x+9}=3-2x\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(3-2x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(3-2x\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-3\right)-\left(3-2x\right)\right]\left[\left(2x-3\right)+\left(3-2x\right)\right]=0\)
\(\Leftrightarrow\left(2x-3-3+2x\right)\left(2x-3+3-2x\right)=0\)
\(\Leftrightarrow\left(4x-6\right)\cdot0=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
2) Ta có: \(\sqrt{x^2-2\cdot\sqrt{2}\cdot x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}-\sqrt{1+2\cdot1\cdot\sqrt{2}+2}\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\left|\sqrt{8}-1\right|-\left|1+\sqrt{2}\right|\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8}-1-1-\sqrt{2}\)
\(\Leftrightarrow\left|x-\sqrt{2}\right|=\sqrt{2}-2\)(*)
Trường hợp 1: \(x\ge\sqrt{2}\)
(*)\(\Leftrightarrow x-\sqrt{2}=\sqrt{2}-2\)
\(\Leftrightarrow x-\sqrt{2}-\sqrt{2}+2=0\)
\(\Leftrightarrow x-2\sqrt{2}+2=0\)
\(\Leftrightarrow x=2\sqrt{2}-2\)(loại)
Trường hợp 2: \(x< \sqrt{2}\)
(*)\(\Leftrightarrow\sqrt{2}-x=\sqrt{2}-2\)
\(\Leftrightarrow\sqrt{2}-x-\sqrt{2}+2=0\)
\(\Leftrightarrow2-x=0\)
hay x=2(loại)
Vậy: S=∅
\(1.4x^2-12x+9=9-12x+4x^2\)
\(0x=0\)
Pt tm với mọi x
